Answer:
it will be 318.5 Kelvin
A parallel-plate capacitor is constructed using adielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00X108V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates.
Answer:
A = 0.188 m²
Explanation:
First we find the distance between the plates by using the formula of electric field intensity:
E = ΔV/d
d = ΔV/E
where,
d = distance between plates = ?
ΔV = Potential Difference = 4 KV = 4000 V
E = Electric Field = 2 x 10⁸ V/m
Therefore,
d = 4000 V/(2 x 10⁸ V/m)
d = 2 x 10⁻⁵ m
Now, we find the Area of Plates by using formula of capacitance:
C = A∈₀∈r/d
where,
C = Capacitance = 0.25 μF = 0.25 x 10⁻⁶ F
A = Area of Plates = ?
∈₀ = Permittivity of free space = 8.85 x 10⁻¹² C/N.m²
∈r = Dielectric Constant = 3
Therefore,
0.25 x 10⁻⁶ F = A(8.85 x 10⁻¹² C/N.m²)(3)/(2 x 10⁻⁵ m)
A = (0.25 x 10⁻⁶ F)(2 x 10⁻⁵ m)/(3)(8.85 x 10⁻¹² C/N.m²)
A = 0.188 m²
PLEASE HELP ME IM TIMED
Answer:
the answer is the core
Explanation:
the core is composed of iron and nickel
Why was basketball invented?
According to the law of universal gravitation, gravity is the force keeping objects in the universe in their relative positions.
true
false
Answer true
Explanation
Jack jumped off a diving board and hit the water at 15.68 m/s downward. How long was he falling for before he hit the water?
Answer:
1.6 s
Explanation:
From the question given above, the following data were:
Velocity (v) = 15.68 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
Thus, we can calculate the time taken for Jack to hit the water by using the following formula:
v = gt
15.68 = 9.8 × t
Divide both side by 9.8
t = 15.68 / 9.8
t = 1.6 s
Therefore, it took Jack 1.6 s to hit the water.
Integrated science please help ASAP!...
Answer:
SewageAgricultural pollutionOilRadioactive substanceRiver dumpingMarine dumpingLittering trash Industrial wasteMining activitiesChemical fertilizersExplanation:
I hope this helps
Jason rides his bicycle for 3 hours at a speed to 40 km/h. What distance does he
travel? -
Answer:
120km
Explanation:
Which of the following charts correctly compares plant and animal cells?
Answer:
Wheres the charts??
Explanation:
6th grade science I mark as brainliest.
Answer:
2m 13[tex]\frac{1}{3}[/tex]s
Explanation:
1.5m = 1s
200m = [tex]\frac{200}{1.5}[/tex] × 1s
= 133[tex]\frac{1}{3}[/tex]s
= 2m 13[tex]\frac{1}{3}[/tex]s
How much power will it take to move a 10 kg mass at an acceleration of 2 m/s² a distance of 10 meters in 5 seconds?
Answer:
100 Watts
Explanation:
These equations are needed to work out the answer:
power= work done/ time takenwork done= force* distanceforce= mass* accelerationforce: 10 kg* 2m/s= 20
work done: 20* 10m=200
power: 200/2=100
plz help me with question ;-;
Answer:
C
Explanation:
Students are asked to design an experiment about Newton’s 2nd Law. One student decides to roll a marble down a ramp into a pile of sand to measure the force impact.
Which variable should she manipulate to best exemplify the relationship explained by this law?
A.She should use a heavier marble, because the marbles will roll at the same rate of acceleration but more mass will produce a larger impact force.
B.She should increase the slope of the ramp by propping it up to higher height, because a steeper ramp will cause a greater rate of acceleration and a larger impact force.
C.She should use a heavier marble, because a bigger marble will accelerate more quickly down the ramp and cause a greater impact force.
D.She should decrease the slope of the ramp, because a ramp with a smaller slope will allow the ball more time to build up speed and cause a greater impact force.
Answer:
A
Explanation:
Trust me I just took it !
The density of an object is dependent upon the object’s mass and ---
Answer:Volume
Explanation:
Density = mass/ Volume
Answer:
Volume
Explanation:
I WILL GIVE BRAINLYIST
Which of the following types of stars had the smallest initial mass? (10 points)
Group of answer choices
Blue main sequence
Neutron star
Red supergiant
White dwarf
Answer:
White dwarf
Explanation:
A fish is 4.7 cm from the front surface of a fish bowl of radius 21 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl? Do not forget the proper sign. (Give your answer in cm.)
______ cm
Where does the fish appear to be when it is 38.9 cm from the front surface of the bowl? (Give your answer in cm.)
______ cm
Answer:
Explanation:
From the information given:
We can properly determine the distance where the fish appear in the air viewing it from in front of the bowl by using the formula:
[tex]\dfrac{n_i}{d_o}+\dfrac{n_2}{d_1}= \dfrac{n_2-n_1}{r}[/tex]
where;
[tex]n_1[/tex] = refractive index in the air; = 1.33 &
[tex]n_2[/tex] = refractive index in water. = 1
[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]
[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{4.7\ cm}[/tex]
[tex]\dfrac{1}{d_i}= - 0.26726 \ cm[/tex]
[tex]d_i =\dfrac{1}{ - 0.26726 \ cm}[/tex]
[tex]\mathbf{d_i }[/tex] = - 3.74 cm
2)
To determine where the fish appear to be when it is 38.9 cm from the front surface of the bowl by using the formula:
[tex]\dfrac{n_2}{d_i}= \dfrac{n_2-n_1}{r}-\dfrac{n_1}{d_o}[/tex]
[tex]\dfrac{1}{d_i}= \dfrac{1-1.33}{-21 \ cm}-\dfrac{1.33}{38.9\ cm}[/tex]
[tex]\dfrac{1}{d_i}=- 0.0184759 \ cm[/tex]
[tex]d_i = \dfrac{1}{- 0.0184759 \ cm}[/tex]
[tex]\mathbf{d_i = }[/tex] -54.12 cm
A car accelerates from rest at a constant acceleration of 25.0 m/s^2. At some point, it then turns off its engine, letting the car decelerate slowly from the force of friction at a constant deceleration of 3 m/s^2 until it is at rest again. The total speed the car moves in this time is 200 meters. What is the minimum time needed for the car to move 200 meters given that it both starts and ends at rest?
Answer:
t = 9.14 s
Explanation:
We first analyze the accelerating motion by applying first equation of motion:
Vf₁ = Vi₁ + a₁t₁
where,
Vf₁ = Final Speed of Car before turning off engine
Vi₁ = Initial Speed of Car = 0 m/s
a₁ = acceleration of car = 25 m/s²
t₁ = time taken in accelerating motion
Therefore,
Vf₁ = 25t₁ ---------- equation (1)
Now, we apply second equation of motion:
s₁ = Vi₁ t₁ + (1/2)a₁t₁²
where,
s₁ = distance covered during accelerating motion
Therefore,
s₁ = (0)t₁ + (1/2)(25)t₁²
s₁ = 12.5 t₁² ----------- equation (2)
Now, we analyze the decelerating motion by applying first equation of motion:
Vf₂ = Vi₂ + a₂t₂
where,
Vf₂ = Final Speed of Car = 0 m/s
Vi₂ = Initial Speed of Car after turning off engine
a₂ = deceleration of car = - 3 m/s²
t₂ = time taken in decelerating motion
Therefore,
Vi₂ = 3t₂ ---------- equation (3)
Now, we apply second equation of motion:
s₂ = Vi₂ t₂ + (1/2)a₂t₂²
where,
s₂ = distance covered during decelerating motion
Therefore,
s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²
s₂ = Vi₂ t₂ - 1.5 t₂²
using equation (3):
s₂ = 3 t₂² - 1.5 t₂²
s₂ = 1.5 t₂² ------------ equation (4)
Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):
Vf₁ = Vi₂
using equation (1) and equation (3):
25 t₁ = 3 t₂
t₁ = 0.12 t₂ ------------ equation (5)
Also, we know that sum of the distances is 200 m:
s₁ + s₂ = 200
using equation (2) and equation (4):
12.5 t₁² + 1.5 t₂² = 200
using equation (5):
12.5 (0.12 t₂²) + 1.5 t₂² = 200
3 t₂² = 200
t₂² = 200/3
t₂ = 8.16 s
substitute this in equation (5):
t₁ = 0.12(8.16 s)
t₁ = 0.97 s
Hence, the minimum time required for this motion is:
t = t₁ + t₂ = 0.97 s + 8.16 s
t = 9.14 s
How do u know that liquid exerts pressure ?
Answer:
All fluids exert pressure like the air inside a tire. The particles of fluids are constantly moving in all directions at random. As the particles move, they keep bumping into each other and into anything else in their path. These collisions cause pressure, and the pressure is exerted equally in all directions.
Explanation:
Hope this helps!! Please consider marking brainliest! Have a good one!!A car travels 3500 m in 200 seconds what is the car speed
Answer:
17.5 m/s
Explanation:
We can calculate the meters per second by dividing the distance by time. 3500 divided by 200 is 17.5, therefore the speed is 17.5 meters per second.
why does a thrown football have both potential and kinetic energy?what is the combined energy called?
i will give brainliest to who ever answers this question in a paraghraph
Answer:
The thrown football has Potential or stored Energy, PE, by virtue of its position in the air and the ability for it to fall by itself. The thrown football also has Kinetic Energy, KE, given that the ball is in motion and requires an equal and opposite amount of energy to stop it. Both the PE and the KE are forms of Mechanical Energy, ME and the Mechanical Energy of the football is equal to the sum of its Potential and Kinetic Energy. That is ME = PE + KE
Explanation:
Potential energy, PE, is the energy that is the held or stored energy of a body such that the body is able to do work without the addition of energy from an external source
A thrown football that has an elevation or height above the floor level has the capacity to come back down and bounce on the floor without the presence of assistance at the topmost height of the football. Therefore, the thrown football has potential energy, given to it by the thrower
Kinetic Energy, KE, is the energy possessed by a moving that comes from the motion of the object
Given that the thrown football is in motion, it posses kinetic energy
Therefore, the thrown football possesses both potential energy and kinetic energy which are forms of mechanical energy ME
The combined energy of the football is therefore called the Mechanical Energy ME of the ball which is the sum of the potential and kinetic energies of the football, given as follows;
The Mechanical Energy of the football = The Potential Energy of the football + The Kinetic Energy of the football
∴ ME = PE + KE
ANSWER QUICK!!
describe two uses for microwave radiation
heat food, warm water
a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.
Answer:
Average speed = ( 2V + V1 + V2)/4
Explanation:
Given that a person covers equal half distance at speed V and remaining half at speed V1 and V2 in equal interval of time .find average speed.
Since the distance is covered at equal intervals of time, and
Speed = distance/time
For the first half distance,
V = distance/t
Cross multiply
Distance = Vt
For the second half distance
(V1 + V2)/2 = distance/t
Distance = t(V1 + V2)/2
The average speed = total distance/ total time.
Average speed = [Vt + t( V1 + V2)/2] ÷ 2t
Average speed = (2Vt + V1t + V2t)/4t
Average speed = t( 2V + V1 + V2)/4t
Time t will cancel out
Average speed = ( 2V + V1 + V2)/4
a block weighing (Fg) 50 N is resting on a steel table (us = 0.74). The minimum force to start this block moving is what N
Answer:37
Explanation:
please help !!!!! I’ll give brainliest !
Answer:
Honestly for me it's a bit too blurry. Sorry luv.:(
Explanation:
Atoms of two different elements must have different
A. Electrical charges
B. Number of neutrons
C. Atomic numbers
D. Energy levels
Explanation:
C. Atomic numbers....
Answer:
atomic numbers
Explanation:
In order for two atoms to be different, they have to have a different number of protons. Protons are represented by the atomic number. Thus, atoms of two DIFFERENT elements must have different atomic numbers.
I took the test and got 100%
Hope this helps!
How much Tension force is required to pull a 1500 kg car (it is being towed) forward with an acceleration of 3 m/s^2 if the friction force on the towed car's tires is pulling backward with a force of 2500 N?
If the pull is done horizontally, then the net force on the car is
∑ F = T - f = (1500 kg) (3 m/s²)
where T is the magnitude of the tension in the towing cable, and f is the friction which points in the opposite direction. Then
T = f + (1500 kg) (3 m/s²)
T = 2500 N + 4500 N
T = 7000 N
Which of the following equations accurately defines acceleration?
Plz help ASAP
Answer:
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
Explanation:
Answer: i think its b
sorry if im wrong
Explanation:
When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the average force was 72.9 N, how much time was it in contact with the floor?
(Unit = s)
Remember: up is +, down is -
Answer:
t = 0.0689 s
Explanation:
Given that,
Mass of a basketball, m = 0.622 kg
Initial velocity, u = 4.23 m/s (downward or negative)
Final velocity, v = 3.85 m/s (up of positive)
Average force, F = 72.9 N
We need to find the time it was in contact with the floor. The force is given by :
[tex]F=ma\\\\F=m\dfrac{v-u}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-(-4.23))}{72.9}\\\\t=0.0689\ s[/tex]
So, the time of contact is 0.0689 s.
Colette launches an air rocket in the upward, positive direction. It launches
with an initial velocity of 25.5 m/s. It accelerates in the downward, negative
direction at a rate of 9.81 m/s2. After 3.5 seconds, what is the magnitude of
the rocket's displacement?
A) 29 meters
B) 31 meters
C) –150 meters
D) 150 meters
Answer:
b
Explanation:
Answer:
The answer is A) 29 meters
Explanation:
I got this question right on the test! :)
A 8.0\,\text {kg}8.0kg8, point, 0, start text, k, g, end text box is released from rest at a height y_0 =0.25\,\text my 0 =0.25my, start subscript, 0, end subscript, equals, 0, point, 25, start text, m, end text on a frictionless ramp. The box slides from the ramp onto a rough horizontal surface. The box slides 2.0\,\text m2.0m2, point, 0, start text, m, end text horizontally until it stops.
Answer:
μ = 0.125
Explanation:
To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.
Let's start working on the ramp
starting point. Highest point of the ramp
Em₀ = U = m h y
final point. Lower part of the ramp, before entering the rough surface
[tex]Em_{f}[/tex] = K = ½ m v²
as they indicate that there is no friction on the ramp
Em₀ = Em_{f}
m g y = ½ m v²
v = [tex]\sqrt{2gy}[/tex]
we calculate
v = √(2 9.8 0.25)
v = 2.21 m / s
in the rough part we use the relationship between work and kinetic energy
W = ΔK = K_{f} -K₀
as it stops the final kinetic energy is zero
W = -K₀
The work is done by the friction force, which opposes the movement
W = - fr x
friction force has the expression
fr = μ N
let's write Newton's second law for the vertical axis
N-W = 0
N = W = m g
we substitute
-μ m g x = - ½ m v²
μ = [tex]\frac{v^{2} }{2 g x}[/tex]
Let's calculate
μ = [tex]\frac{2.21^{2}}{2\ 9.8\ 2.0}[/tex]
μ = 0.125
A 8.0 kg box is released from rest at a height y0 = 0.25 m on a frictionless ramp. The box slides from the ramp onto a rough horizontal surface. The box slides 2.0 m horizontally until it stops.
What is the friction coefficient of the horizontal surface?
Answer: 0.125
A 0.050 kg ball starts from rest at some unknown height on a toy roller coaster.
At a later time, it travels through the top of a loop at 2 m/s and a height of 0.40 m.
Since this track is frictionless, what was the starting height of the ball?
Answer:
The starting height of the ball is approximately 0.604 m
Explanation:
The given parameters are;
The mass of the the ball = 0.050
The speed with which it travels through the top loop = 2 m/s
The given height at which the ball moves at 2 m/s = 0.40 m
Therefore, we have;
1/2·m·v² = m·g·h
1/2·v² = g·h
h = 1/2·v²/g = 1/2 × 2²/9.81 ≈ 0.204
The additional height = h = 0.204 m
Therefore;
The starting height of the ball ≈ The given height at which the ball moves at 2 m/s + h
The starting height of the ball ≈ 0.40 + 0.204 = 0.604 m
The starting height of the ball ≈ 0.604 m.
When any object in rest, then potential energy is present. But when object is in motion then object have kinetic energy.
Starting height of the ball is 0.604 m.
We know that, when any object is start from rest, then potential energy is converted into kinetic energy.
[tex]\frac{1}{2}mv^{2} =mgh[/tex]
Where m is mass of object, g is gravitational acceleration , h is height and v is velocity of object. (value of g = 9.81 m/ second square)
from above equation,
we get, extra height [tex]h=\frac{v^{2} }{2g} \\\\h=\frac{4}{2*9.81}\\\\h=0.204[/tex] meter
The starting height of the ball will be sum of the height at which ball moves 2 m/s and extra height.
Starting height = 0.40 + 0.204 = 0.604 meter.
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